Question:
The rest energy of an electron is 0.511 MeV. The electron is accelerated from rest to a velocity of 0.5c. The change in its energy will be:
The rest energy of an electron is 0.511 MeV. The electron is accelerated from rest to a velocity of 0.5c. The change in its energy will be:
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Relativistic energy increases with velocity, and the energy change can be calculated using the Lorentz factor.
Updated On: Apr 22, 2026
- 0.026 MeV
- 0.051 MeV
- 0.079 MeV
- 0.105 MeV
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The Correct Option is C
Solution and Explanation
Step 1: Energy change due to relativistic motion.
The total energy \( E \) of a moving particle is given by: \[ E = \gamma m c^2 \] where \( \gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} \) is the Lorentz factor, \( v \) is the velocity of the electron, and \( c \) is the speed of light.
Step 2: Calculate the change in energy.
For an electron accelerated to \( 0.5c \), we calculate \( \gamma \): \[ \gamma = \frac{1}{\sqrt{1 - \frac{(0.5c)^2}{c^2}}} = \frac{1}{\sqrt{1 - 0.25}} = \frac{1}{\sqrt{0.75}} \approx 1.1547 \] The total energy is: \[ E = \gamma \times 0.511 \, \text{MeV} = 1.1547 \times 0.511 = 0.590 \, \text{MeV} \] Thus, the change in energy is: \[ \Delta E = E - E_0 = 0.590 \, \text{MeV} - 0.511 \, \text{MeV} = 0.079 \, \text{MeV} \]
The total energy \( E \) of a moving particle is given by: \[ E = \gamma m c^2 \] where \( \gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} \) is the Lorentz factor, \( v \) is the velocity of the electron, and \( c \) is the speed of light.
Step 2: Calculate the change in energy.
For an electron accelerated to \( 0.5c \), we calculate \( \gamma \): \[ \gamma = \frac{1}{\sqrt{1 - \frac{(0.5c)^2}{c^2}}} = \frac{1}{\sqrt{1 - 0.25}} = \frac{1}{\sqrt{0.75}} \approx 1.1547 \] The total energy is: \[ E = \gamma \times 0.511 \, \text{MeV} = 1.1547 \times 0.511 = 0.590 \, \text{MeV} \] Thus, the change in energy is: \[ \Delta E = E - E_0 = 0.590 \, \text{MeV} - 0.511 \, \text{MeV} = 0.079 \, \text{MeV} \]
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